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LenKa [72]
3 years ago
7

−6 < x + 1 ≤ 1 is what kind of linear inequality.

Mathematics
1 answer:
Darya [45]3 years ago
3 0
-6 < x + 1 ≤ 1
<u>-1        -1    -1</u>
-7 < x ≤ 0
-7 < x or x ≤ 0
x > -7 or x ≤ 0
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If the volume of a cube-shaped compost bin is 27 cubic feet. What is the edge length of the compost bin?
Alex787 [66]

Cube = all sides same length, which equals A*A*A = 27 Which is the same as A^3 = 27 Therefore,the cube root is +/-3, but as it is length, must be positive, so is +3

3 0
3 years ago
This is 10 points plz help ( NO LINKS )
monitta

Answer:

1500 pounds

Step-by-step explanation:

math lol

8 0
3 years ago
Solve for the missing variables in the following equations
exis [7]
75 = 14y + 5

14y = 75 - 5 = 70

14y = 70

y = \frac{70}{14} =  \frac{10}{2}

y = 5
6 0
3 years ago
Modeled with
scZoUnD [109]

x² - 3x - 10

x² = x * x

10 can be factored:

1 x 10 ⇒⇒⇒ difference between factors = 9

2 x 5   ⇒⇒⇒ difference between factors = 3  ⇒ (The correct pair)

(x-5)(x+2) = x(x+2) - 5(x+2)

                = x² + 2x - 5x - 10

                = x² - 3x - 10

x - 5 = 0  ⇒⇒⇒ x = 5

  x + 2 = 0  ⇒⇒⇒ x = -2

8 0
3 years ago
A very large data set (N &gt; 10,000) has a mean value of 1.65 units and a standard deviation of 72.26 units. Determine the rang
Romashka [77]

Answer:

z=-0.674

And if we solve for a we got

a=1.65 -0.674*72.26=-47.05

z=0.674

And if we solve for a we got

a=1.65 +0.674*72.26=50.35

So then the limits where 50% of the data lies are -47.05 and 50.35

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that variable of interest of a population, and for this case we know the distribution for X is given by:

X \sim N(1.65,72.26)  

Where \mu=1.65 and \sigma=72.26

For this case we want the limits for the 50% of the values.

So on the tails of the distribution we need the other 50% of the data, and on ach tail we need to have 25% since the distribution is symmetric.

Lower tail

For this part we want to find a value a, such that we satisfy this condition:

P(X>a)=0.75   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=-0.674

And if we solve for a we got

a=1.65 -0.674*72.26=-47.05

So the value of height that separates the bottom 25% of data from the top 75% is -47.05.  

Upper tail

For this part we want to find a value a, such that we satisfy this condition:

P(X>a)=0.25   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.75 of the area on the left and 0.25 of the area on the right it's z=0.674. On this case P(Z<0.674)=0.75 and P(z>0.674)=0.25

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=0.674

And if we solve for a we got

a=1.65 +0.674*72.26=50.35

So the value of height that separates the bottom 75% of data from the top 25% is 50.35.  

So then the limits where 50% of the data lies are -47.05 and 50.35

6 0
3 years ago
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